Consider a circle with a radius of length \(r=7.2\) meters. A central angle, in standard position, subtends an arc of length \(14.4\) meters.
Find \(\theta\) in radians.
Solution
Remember, the radian measure of an angle is the length of the arc on the unit circle subtended by the angle. The circle shown is not a unit circle (its radius is 7.2), but we can add a unit circle into the drawing.
(Not drawn to scale.)
\[\frac{\theta}{14.4}~=~\frac{1}{7.2}\]
\[\theta~=~\frac{14.4}{7.2}~=~2\]
If you refer to the wikipedia article on circular arcs, you’ll see a formula:
\[L=\theta r\]
where \(L\) is an arc length on a circle of radius \(r\) and with central angle \(\theta\) (pronounced “theta”). In our case, we were given \(L=14.4\) and \(r=7.2\), so we can rearrange the formula, solving for \(\theta\).
\[\theta = \frac{L}{r} = \frac{14.4}{7.2}\]
Question
Consider a circle with a radius of length \(r=9.6\) meters. A central angle, in standard position, subtends an arc of length \(45.6\) meters.
Find \(\theta\) in radians.
Solution
Remember, the radian measure of an angle is the length of the arc on the unit circle subtended by the angle. The circle shown is not a unit circle (its radius is 9.6), but we can add a unit circle into the drawing.
(Not drawn to scale.)
\[\frac{\theta}{45.6}~=~\frac{1}{9.6}\]
\[\theta~=~\frac{45.6}{9.6}~=~4.75\]
If you refer to the wikipedia article on circular arcs, you’ll see a formula:
\[L=\theta r\]
where \(L\) is an arc length on a circle of radius \(r\) and with central angle \(\theta\) (pronounced “theta”). In our case, we were given \(L=45.6\) and \(r=9.6\), so we can rearrange the formula, solving for \(\theta\).
\[\theta = \frac{L}{r} = \frac{45.6}{9.6}\]
Question
Consider a circle with a radius of length \(r=5.1\) meters. A central angle, in standard position, subtends an arc of length \(10.2\) meters.
Find \(\theta\) in radians.
Solution
Remember, the radian measure of an angle is the length of the arc on the unit circle subtended by the angle. The circle shown is not a unit circle (its radius is 5.1), but we can add a unit circle into the drawing.
(Not drawn to scale.)
\[\frac{\theta}{10.2}~=~\frac{1}{5.1}\]
\[\theta~=~\frac{10.2}{5.1}~=~2\]
If you refer to the wikipedia article on circular arcs, you’ll see a formula:
\[L=\theta r\]
where \(L\) is an arc length on a circle of radius \(r\) and with central angle \(\theta\) (pronounced “theta”). In our case, we were given \(L=10.2\) and \(r=5.1\), so we can rearrange the formula, solving for \(\theta\).
\[\theta = \frac{L}{r} = \frac{10.2}{5.1}\]
Question
Consider a circle with a radius of length \(r=6.9\) meters. A central angle, in standard position, subtends an arc of length \(13.8\) meters.
Find \(\theta\) in radians.
Solution
Remember, the radian measure of an angle is the length of the arc on the unit circle subtended by the angle. The circle shown is not a unit circle (its radius is 6.9), but we can add a unit circle into the drawing.
(Not drawn to scale.)
\[\frac{\theta}{13.8}~=~\frac{1}{6.9}\]
\[\theta~=~\frac{13.8}{6.9}~=~2\]
If you refer to the wikipedia article on circular arcs, you’ll see a formula:
\[L=\theta r\]
where \(L\) is an arc length on a circle of radius \(r\) and with central angle \(\theta\) (pronounced “theta”). In our case, we were given \(L=13.8\) and \(r=6.9\), so we can rearrange the formula, solving for \(\theta\).
\[\theta = \frac{L}{r} = \frac{13.8}{6.9}\]
Question
Consider a circle with a radius of length \(r=9\) meters. A central angle, in standard position, subtends an arc of length \(36.9\) meters.
Find \(\theta\) in radians.
Solution
Remember, the radian measure of an angle is the length of the arc on the unit circle subtended by the angle. The circle shown is not a unit circle (its radius is 9), but we can add a unit circle into the drawing.
(Not drawn to scale.)
\[\frac{\theta}{36.9}~=~\frac{1}{9}\]
\[\theta~=~\frac{36.9}{9}~=~4.1\]
If you refer to the wikipedia article on circular arcs, you’ll see a formula:
\[L=\theta r\]
where \(L\) is an arc length on a circle of radius \(r\) and with central angle \(\theta\) (pronounced “theta”). In our case, we were given \(L=36.9\) and \(r=9\), so we can rearrange the formula, solving for \(\theta\).
\[\theta = \frac{L}{r} = \frac{36.9}{9}\]
Question
Which graph shows an angle measuring \(\frac{4\pi}{9}\) radians in standard position?
A
B
C
D
E
F
G
H
Solution
The given angle is \(\frac{4\pi}{9}\) radians. The angle is positive, so the angle rotates counterclockwise. I prefer to think of this as \(4\) times \(\frac{\pi}{9}\), so then I imagine the top semicircle (and bottom semicircle) broken up into 9 equally-spaced wedges.
Question
Which graph shows an angle measuring \(\frac{11\pi}{7}\) radians in standard position?
A
B
C
D
E
F
G
H
Solution
The given angle is \(\frac{11\pi}{7}\) radians. The angle is positive, so the angle rotates counterclockwise. I prefer to think of this as \(11\) times \(\frac{\pi}{7}\), so then I imagine the top semicircle (and bottom semicircle) broken up into 7 equally-spaced wedges.
Question
Which graph shows an angle measuring \(\frac{10\pi}{11}\) radians in standard position?
A
B
C
D
E
F
G
H
Solution
The given angle is \(\frac{10\pi}{11}\) radians. The angle is positive, so the angle rotates counterclockwise. I prefer to think of this as \(10\) times \(\frac{\pi}{11}\), so then I imagine the top semicircle (and bottom semicircle) broken up into 11 equally-spaced wedges.
Question
Which graph shows an angle measuring \(\frac{23\pi}{12}\) radians in standard position?
A
B
C
D
E
F
G
H
Solution
The given angle is \(\frac{23\pi}{12}\) radians. The angle is positive, so the angle rotates counterclockwise. I prefer to think of this as \(23\) times \(\frac{\pi}{12}\), so then I imagine the top semicircle (and bottom semicircle) broken up into 12 equally-spaced wedges.
Question
Which graph shows an angle measuring \(\frac{\pi}{7}\) radians in standard position?
A
B
C
D
E
F
G
H
Solution
The given angle is \(\frac{\pi}{7}\) radians. The angle is positive, so the angle rotates counterclockwise. I prefer to think of this as \(1\) times \(\frac{\pi}{7}\), so then I imagine the top semicircle (and bottom semicircle) broken up into 7 equally-spaced wedges.
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\sin(0.273)~=~\frac{x}{94.6}\]
So,
\[x~=~94.6\cdot \sin(0.273)\]
\[x=25.5061981\]
Express with 3 significant figures.
\[x=25.5\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\sin(1.04)~=~\frac{x}{81.4}\]
So,
\[x~=~81.4\cdot \sin(1.04)\]
\[x=70.1997041\]
Express with 3 significant figures.
\[x=70.2\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\sin(0.78)~=~\frac{x}{37.7}\]
So,
\[x~=~37.7\cdot \sin(0.78)\]
\[x=26.5136341\]
Express with 3 significant figures.
\[x=26.5\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\sin(0.956)~=~\frac{x}{22.9}\]
So,
\[x~=~22.9\cdot \sin(0.956)\]
\[x=18.7068025\]
Express with 3 significant figures.
\[x=18.7\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\sin(0.351)~=~\frac{x}{93.9}\]
So,
\[x~=~93.9\cdot \sin(0.351)\]
\[x=32.2862951\]
Express with 3 significant figures.
\[x=32.3\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\sin(0.658)~=~\frac{58.4}{x}\]
So,
\[x~=~\frac{58.4}{\sin(0.658)}\]
\[x=95.4972956\]
Express with 3 significant figures.
\[x=95.5\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\sin(0.786)~=~\frac{64.9}{x}\]
So,
\[x~=~\frac{64.9}{\sin(0.786)}\]
\[x=91.727272\]
Express with 3 significant figures.
\[x=91.7\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\sin(0.391)~=~\frac{16.3}{x}\]
So,
\[x~=~\frac{16.3}{\sin(0.391)}\]
\[x=42.7694523\]
Express with 3 significant figures.
\[x=42.8\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\sin(0.65)~=~\frac{25.3}{x}\]
So,
\[x~=~\frac{25.3}{\sin(0.65)}\]
\[x=41.8053012\]
Express with 3 significant figures.
\[x=41.8\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\sin(0.894)~=~\frac{15.8}{x}\]
So,
\[x~=~\frac{15.8}{\sin(0.894)}\]
\[x=20.267241\]
Express with 3 significant figures.
\[x=20.3\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\sin(x)~=~\frac{24.7}{29.2}\]
So,
\[x~=~\arcsin\left(\frac{24.7}{29.2}\right)\]
\[x=1.0082324\]
Express with 3 significant figures.
\[x=1.01\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\sin(x)~=~\frac{45.7}{54}\]
So,
\[x~=~\arcsin\left(\frac{45.7}{54}\right)\]
\[x=1.0089939\]
Express with 3 significant figures.
\[x=1.01\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\sin(x)~=~\frac{49.2}{53.4}\]
So,
\[x~=~\arcsin\left(\frac{49.2}{53.4}\right)\]
\[x=1.1715348\]
Express with 3 significant figures.
\[x=1.17\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\sin(x)~=~\frac{24}{27.5}\]
So,
\[x~=~\arcsin\left(\frac{24}{27.5}\right)\]
\[x=1.060761\]
Express with 3 significant figures.
\[x=1.06\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\sin(x)~=~\frac{55.2}{71.7}\]
So,
\[x~=~\arcsin\left(\frac{55.2}{71.7}\right)\]
\[x=0.8786444\]
Express with 3 significant figures.
\[x=0.879\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\cos(0.286)~=~\frac{x}{94.5}\]
So,
\[x~=~94.5\cdot \cos(0.286)\]
\[x=90.6614115\]
Express with 3 significant figures.
\[x=90.7\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\cos(0.716)~=~\frac{x}{56.5}\]
So,
\[x~=~56.5\cdot \cos(0.716)\]
\[x=42.6257044\]
Express with 3 significant figures.
\[x=42.6\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\cos(0.842)~=~\frac{x}{71.9}\]
So,
\[x~=~71.9\cdot \cos(0.842)\]
\[x=47.8834016\]
Express with 3 significant figures.
\[x=47.9\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\cos(0.756)~=~\frac{x}{97.3}\]
So,
\[x~=~97.3\cdot \cos(0.756)\]
\[x=70.7941071\]
Express with 3 significant figures.
\[x=70.8\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\cos(1.05)~=~\frac{x}{57.9}\]
So,
\[x~=~57.9\cdot \cos(1.05)\]
\[x=28.8093637\]
Express with 3 significant figures.
\[x=28.8\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\cos(0.27)~=~\frac{75.5}{x}\]
So,
\[x~=~\frac{75.5}{\cos(0.27)}\]
\[x=78.3381199\]
Express with 3 significant figures.
\[x=78.3\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\cos(0.938)~=~\frac{58.8}{x}\]
So,
\[x~=~\frac{58.8}{\cos(0.938)}\]
\[x=99.4247634\]
Express with 3 significant figures.
\[x=99.4\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\cos(0.627)~=~\frac{21.5}{x}\]
So,
\[x~=~\frac{21.5}{\cos(0.627)}\]
\[x=26.5500505\]
Express with 3 significant figures.
\[x=26.6\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\cos(0.897)~=~\frac{12.6}{x}\]
So,
\[x~=~\frac{12.6}{\cos(0.897)}\]
\[x=20.1936944\]
Express with 3 significant figures.
\[x=20.2\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\cos(0.65)~=~\frac{40.3}{x}\]
So,
\[x~=~\frac{40.3}{\cos(0.65)}\]
\[x=50.6228114\]
Express with 3 significant figures.
\[x=50.6\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\cos(x)~=~\frac{39.6}{67.7}\]
So,
\[x~=~\arccos\left(\frac{39.6}{67.7}\right)\]
\[x=0.9459982\]
Express with 3 significant figures.
\[x=0.946\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\cos(x)~=~\frac{91.6}{95.1}\]
So,
\[x~=~\arccos\left(\frac{91.6}{95.1}\right)\]
\[x=0.2721446\]
Express with 3 significant figures.
\[x=0.272\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\cos(x)~=~\frac{22}{37.7}\]
So,
\[x~=~\arccos\left(\frac{22}{37.7}\right)\]
\[x=0.9476976\]
Express with 3 significant figures.
\[x=0.948\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\cos(x)~=~\frac{68}{78.1}\]
So,
\[x~=~\arccos\left(\frac{68}{78.1}\right)\]
\[x=0.514216\]
Express with 3 significant figures.
\[x=0.514\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\cos(x)~=~\frac{47.7}{52.4}\]
So,
\[x~=~\arccos\left(\frac{47.7}{52.4}\right)\]
\[x=0.4267752\]
Express with 3 significant figures.
\[x=0.427\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\tan(0.961)~=~\frac{x}{46.7}\]
So,
\[x~=~46.7\cdot \tan(0.961)\]
\[x=66.8464755\]
Express with 3 significant figures.
\[x=66.8\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\tan(0.342)~=~\frac{x}{57.5}\]
So,
\[x~=~57.5\cdot \tan(0.342)\]
\[x=20.4693522\]
Express with 3 significant figures.
\[x=20.5\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\tan(0.841)~=~\frac{x}{31.9}\]
So,
\[x~=~31.9\cdot \tan(0.841)\]
\[x=35.6603558\]
Express with 3 significant figures.
\[x=35.7\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\tan(0.764)~=~\frac{x}{54.6}\]
So,
\[x~=~54.6\cdot \tan(0.764)\]
\[x=52.3119318\]
Express with 3 significant figures.
\[x=52.3\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\tan(0.416)~=~\frac{x}{77.5}\]
So,
\[x~=~77.5\cdot \tan(0.416)\]
\[x=34.238211\]
Express with 3 significant figures.
\[x=34.2\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\tan(0.629)~=~\frac{45.5}{x}\]
So,
\[x~=~\frac{45.5}{\tan(0.629)}\]
\[x=62.5357144\]
Express with 3 significant figures.
\[x=62.5\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\tan(1.24)~=~\frac{41.2}{x}\]
So,
\[x~=~\frac{41.2}{\tan(1.24)}\]
\[x=14.1486924\]
Express with 3 significant figures.
\[x=14.1\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\tan(1.01)~=~\frac{46.5}{x}\]
So,
\[x~=~\frac{46.5}{\tan(1.01)}\]
\[x=29.2047633\]
Express with 3 significant figures.
\[x=29.2\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\tan(0.918)~=~\frac{15.2}{x}\]
So,
\[x~=~\frac{15.2}{\tan(0.918)}\]
\[x=11.6223177\]
Express with 3 significant figures.
\[x=11.6\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\tan(0.827)~=~\frac{22.3}{x}\]
So,
\[x~=~\frac{22.3}{\tan(0.827)}\]
\[x=20.5176776\]
Express with 3 significant figures.
\[x=20.5\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\tan(x)~=~\frac{39.3}{67.1}\]
So,
\[x~=~\arctan\left(\frac{39.3}{67.1}\right)\]
\[x=0.5298332\]
Express with 3 significant figures.
\[x=0.530\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\tan(x)~=~\frac{34.7}{38.4}\]
So,
\[x~=~\arctan\left(\frac{34.7}{38.4}\right)\]
\[x=0.7348257\]
Express with 3 significant figures.
\[x=0.735\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\tan(x)~=~\frac{82.1}{28.2}\]
So,
\[x~=~\arctan\left(\frac{82.1}{28.2}\right)\]
\[x=1.2399386\]
Express with 3 significant figures.
\[x=1.24\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\tan(x)~=~\frac{26.5}{87.6}\]
So,
\[x~=~\arctan\left(\frac{26.5}{87.6}\right)\]
\[x=0.2937592\]
Express with 3 significant figures.
\[x=0.294\]
Question
Find the length or angle measure marked by \(x\). Please express your answer with 3 significant figures.
Solution
\[\tan(x)~=~\frac{33.8}{45.3}\]
So,
\[x~=~\arctan\left(\frac{33.8}{45.3}\right)\]
\[x=0.6410241\]
Express with 3 significant figures.
\[x=0.641\]
Question
An angle, \(\theta\), is shown below in standard position as a blue spiraling arrow. A radius (of length 1) is drawn at that angle. Where the radius connects to the circumference of the unit circle has Cartesian coordinates \((-0.761, 0.649)\).
Calculate the trigonometric ratios of \(\theta\).
The sine: \(\sin(\theta) =\)
The cosine: \(\cos(\theta)=\)
The tangent: \(\tan(\theta)=\)
Solution
I’ve listed all 6 trigonometric ratios below. In this class, we will only concern ourselves with sine, cosine, and tangent.
function
expression
number
sine
\(y\)
0.649
cosine
\(x\)
-0.761
tangent
\(\frac{y}{x}\)
-0.853
cosecant
\(\frac{1}{y}\)
1.541
secant
\(\frac{1}{x}\)
-1.314
cotangent
\(\frac{x}{y}\)
-1.173
Question
An angle, \(\theta\), is shown below in standard position as a blue spiraling arrow. A radius (of length 1) is drawn at that angle. Where the radius connects to the circumference of the unit circle has Cartesian coordinates \((0.743, 0.669)\).
Calculate the trigonometric ratios of \(\theta\).
The sine: \(\sin(\theta) =\)
The cosine: \(\cos(\theta)=\)
The tangent: \(\tan(\theta)=\)
Solution
I’ve listed all 6 trigonometric ratios below. In this class, we will only concern ourselves with sine, cosine, and tangent.
function
expression
number
sine
\(y\)
0.669
cosine
\(x\)
0.743
tangent
\(\frac{y}{x}\)
0.9
cosecant
\(\frac{1}{y}\)
1.495
secant
\(\frac{1}{x}\)
1.346
cotangent
\(\frac{x}{y}\)
1.111
Question
An angle, \(\theta\), is shown below in standard position as a blue spiraling arrow. A radius (of length 1) is drawn at that angle. Where the radius connects to the circumference of the unit circle has Cartesian coordinates \((0.468, 0.884)\).
Calculate the trigonometric ratios of \(\theta\).
The sine: \(\sin(\theta) =\)
The cosine: \(\cos(\theta)=\)
The tangent: \(\tan(\theta)=\)
Solution
I’ve listed all 6 trigonometric ratios below. In this class, we will only concern ourselves with sine, cosine, and tangent.
function
expression
number
sine
\(y\)
0.884
cosine
\(x\)
0.468
tangent
\(\frac{y}{x}\)
1.889
cosecant
\(\frac{1}{y}\)
1.131
secant
\(\frac{1}{x}\)
2.137
cotangent
\(\frac{x}{y}\)
0.529
Question
An angle, \(\theta\), is shown below in standard position as a blue spiraling arrow. A radius (of length 1) is drawn at that angle. Where the radius connects to the circumference of the unit circle has Cartesian coordinates \((-0.556, -0.831)\).
Calculate the trigonometric ratios of \(\theta\).
The sine: \(\sin(\theta) =\)
The cosine: \(\cos(\theta)=\)
The tangent: \(\tan(\theta)=\)
Solution
I’ve listed all 6 trigonometric ratios below. In this class, we will only concern ourselves with sine, cosine, and tangent.
function
expression
number
sine
\(y\)
-0.831
cosine
\(x\)
-0.556
tangent
\(\frac{y}{x}\)
1.495
cosecant
\(\frac{1}{y}\)
-1.203
secant
\(\frac{1}{x}\)
-1.799
cotangent
\(\frac{x}{y}\)
0.669
Question
An angle, \(\theta\), is shown below in standard position as a blue spiraling arrow. A radius (of length 1) is drawn at that angle. Where the radius connects to the circumference of the unit circle has Cartesian coordinates \((-0.822, 0.569)\).
Calculate the trigonometric ratios of \(\theta\).
The sine: \(\sin(\theta) =\)
The cosine: \(\cos(\theta)=\)
The tangent: \(\tan(\theta)=\)
Solution
I’ve listed all 6 trigonometric ratios below. In this class, we will only concern ourselves with sine, cosine, and tangent.
function
expression
number
sine
\(y\)
0.569
cosine
\(x\)
-0.822
tangent
\(\frac{y}{x}\)
-0.692
cosecant
\(\frac{1}{y}\)
1.757
secant
\(\frac{1}{x}\)
-1.217
cotangent
\(\frac{x}{y}\)
-1.445
Question
Given:
\[\cos(\theta)~=~-0.428\]\[\theta~\text{ is in quadrant III}\]
Find \(\sin(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\cos(\theta)\) represents the \(x\) coordinate on the unit circle at an angle of \(\theta\), and \(\sin(\theta)\) represents the \(y\) coordinate. Draw a sketch. We know the point of interest is in quadrant III.
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment at \(x=-0.428\). I’ve labelled the lengths using positive numbers. I will later consider the signs (positive or negative) of \(x\) and \(y\) based on the quadrant.
Use Pythagorean Theorem.
\[(-0.428)^2+|y|^2~=~1^2\]
The square of a real number always equals the square of the absolute value of the same real number. Also, \(1^2=1\).
\[0.428^2+y^2~=~1\]
Subtract \(0.428^2\) from both sides.
\[y^2~=~1-0.428^2\]
Square root both sides.
\[|y|~=~ \sqrt{(1-0.428^2)}\]
We know the quadrant is quadrant III, so \(y\) is negative.
\[y~=~-0.9037787\]
So,
\[\sin(\theta)~=~-0.9037787\]
Question
Given:
\[\cos(\theta)~=~-0.333\]\[\theta~\text{ is in quadrant II}\]
Find \(\sin(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\cos(\theta)\) represents the \(x\) coordinate on the unit circle at an angle of \(\theta\), and \(\sin(\theta)\) represents the \(y\) coordinate. Draw a sketch. We know the point of interest is in quadrant II.
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment at \(x=-0.333\). I’ve labelled the lengths using positive numbers. I will later consider the signs (positive or negative) of \(x\) and \(y\) based on the quadrant.
Use Pythagorean Theorem.
\[(-0.333)^2+|y|^2~=~1^2\]
The square of a real number always equals the square of the absolute value of the same real number. Also, \(1^2=1\).
\[0.333^2+y^2~=~1\]
Subtract \(0.333^2\) from both sides.
\[y^2~=~1-0.333^2\]
Square root both sides.
\[|y|~=~ \sqrt{(1-0.333^2)}\]
We know the quadrant is quadrant II, so \(y\) is positive.
\[y~=~0.9429268\]
So,
\[\sin(\theta)~=~0.9429268\]
Question
Given:
\[\cos(\theta)~=~-0.615\]\[\theta~\text{ is in quadrant III}\]
Find \(\sin(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\cos(\theta)\) represents the \(x\) coordinate on the unit circle at an angle of \(\theta\), and \(\sin(\theta)\) represents the \(y\) coordinate. Draw a sketch. We know the point of interest is in quadrant III.
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment at \(x=-0.615\). I’ve labelled the lengths using positive numbers. I will later consider the signs (positive or negative) of \(x\) and \(y\) based on the quadrant.
Use Pythagorean Theorem.
\[(-0.615)^2+|y|^2~=~1^2\]
The square of a real number always equals the square of the absolute value of the same real number. Also, \(1^2=1\).
\[0.615^2+y^2~=~1\]
Subtract \(0.615^2\) from both sides.
\[y^2~=~1-0.615^2\]
Square root both sides.
\[|y|~=~ \sqrt{(1-0.615^2)}\]
We know the quadrant is quadrant III, so \(y\) is negative.
\[y~=~-0.7885271\]
So,
\[\sin(\theta)~=~-0.7885271\]
Question
Given:
\[\cos(\theta)~=~0.801\]\[\theta~\text{ is in quadrant I}\]
Find \(\sin(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\cos(\theta)\) represents the \(x\) coordinate on the unit circle at an angle of \(\theta\), and \(\sin(\theta)\) represents the \(y\) coordinate. Draw a sketch. We know the point of interest is in quadrant I.
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment at \(x=0.801\). I’ve labelled the lengths using positive numbers. I will later consider the signs (positive or negative) of \(x\) and \(y\) based on the quadrant.
Use Pythagorean Theorem.
\[(0.801)^2+|y|^2~=~1^2\]
The square of a real number always equals the square of the absolute value of the same real number. Also, \(1^2=1\).
\[0.801^2+y^2~=~1\]
Subtract \(0.801^2\) from both sides.
\[y^2~=~1-0.801^2\]
Square root both sides.
\[|y|~=~ \sqrt{(1-0.801^2)}\]
We know the quadrant is quadrant I, so \(y\) is positive.
\[y~=~0.5986643\]
So,
\[\sin(\theta)~=~0.5986643\]
Question
Given:
\[\cos(\theta)~=~-0.168\]\[\theta~\text{ is in quadrant II}\]
Find \(\sin(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\cos(\theta)\) represents the \(x\) coordinate on the unit circle at an angle of \(\theta\), and \(\sin(\theta)\) represents the \(y\) coordinate. Draw a sketch. We know the point of interest is in quadrant II.
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment at \(x=-0.168\). I’ve labelled the lengths using positive numbers. I will later consider the signs (positive or negative) of \(x\) and \(y\) based on the quadrant.
Use Pythagorean Theorem.
\[(-0.168)^2+|y|^2~=~1^2\]
The square of a real number always equals the square of the absolute value of the same real number. Also, \(1^2=1\).
\[0.168^2+y^2~=~1\]
Subtract \(0.168^2\) from both sides.
\[y^2~=~1-0.168^2\]
Square root both sides.
\[|y|~=~ \sqrt{(1-0.168^2)}\]
We know the quadrant is quadrant II, so \(y\) is positive.
\[y~=~0.985787\]
So,
\[\sin(\theta)~=~0.985787\]
Question
Given:
\[\tan(\theta)~=~-3.8\]\[\theta~\text{ is in quadrant IV}\]
Find \(\sin(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\tan(\theta)\) represents the slope of the segment connecting \((0,0)\) to \((\cos(\theta),\sin(\theta))\). Draw a sketch. We know the point of interest is in quadrant IV. Draw the unit circle centered at the origin and the line \(y=-3.8x\).
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment from the \(x\)-axis to the point. In order for the slope to equal \(\frac{y}{x}\), we can enforce that \(|y|=|-3.8x|\). Notice I am labeling the lengths with positive numbers; I will take care of signs (whether \(x\) or \(y\) is positive or negative), after using Pythagorean Theorem, by considering the quadrant.
Use Pythagorean Theorem.
\[x^2+(3.8x)^2~=~1^2\]
Distribute the power of 2 across the two factors in the parentheses. Also, \(1^2=1\).
\[x^2+14.44x^2~=~1\]
Combine like terms. \(1x^2+14.44x^2\equiv (1+14.44)x^2 \equiv 15.44x^2\)
\[15.44x^2~=~1\]
Divide both sides by 15.44.
\[x^2~=~\frac{1}{15.44}\]
Square root both sides.
\[|x|~=~ \sqrt{\frac{1}{15.44}}\]
We know the quadrant is quadrant IV, so \(x\) is positive.
\[x~=~ 0.2544933\]
We know that \(y=-3.8x\).
\[y~=~(-3.8)(0.2544933)\]
\[y~=~-0.9670745\]
So,
\[\sin(\theta)~=~-0.9670745\]
Question
Given:
\[\tan(\theta)~=~2.5\]\[\theta~\text{ is in quadrant I}\]
Find \(\sin(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\tan(\theta)\) represents the slope of the segment connecting \((0,0)\) to \((\cos(\theta),\sin(\theta))\). Draw a sketch. We know the point of interest is in quadrant I. Draw the unit circle centered at the origin and the line \(y=2.5x\).
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment from the \(x\)-axis to the point. In order for the slope to equal \(\frac{y}{x}\), we can enforce that \(|y|=|2.5x|\). Notice I am labeling the lengths with positive numbers; I will take care of signs (whether \(x\) or \(y\) is positive or negative), after using Pythagorean Theorem, by considering the quadrant.
Use Pythagorean Theorem.
\[x^2+(2.5x)^2~=~1^2\]
Distribute the power of 2 across the two factors in the parentheses. Also, \(1^2=1\).
\[x^2+6.25x^2~=~1\]
Combine like terms. \(1x^2+6.25x^2\equiv (1+6.25)x^2 \equiv 7.25x^2\)
\[7.25x^2~=~1\]
Divide both sides by 7.25.
\[x^2~=~\frac{1}{7.25}\]
Square root both sides.
\[|x|~=~ \sqrt{\frac{1}{7.25}}\]
We know the quadrant is quadrant I, so \(x\) is positive.
\[x~=~ 0.3713907\]
We know that \(y=2.5x\).
\[y~=~(2.5)(0.3713907)\]
\[y~=~0.9284767\]
So,
\[\sin(\theta)~=~0.9284767\]
Question
Given:
\[\tan(\theta)~=~0.823\]\[\theta~\text{ is in quadrant III}\]
Find \(\sin(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\tan(\theta)\) represents the slope of the segment connecting \((0,0)\) to \((\cos(\theta),\sin(\theta))\). Draw a sketch. We know the point of interest is in quadrant III. Draw the unit circle centered at the origin and the line \(y=0.823x\).
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment from the \(x\)-axis to the point. In order for the slope to equal \(\frac{y}{x}\), we can enforce that \(|y|=|0.823x|\). Notice I am labeling the lengths with positive numbers; I will take care of signs (whether \(x\) or \(y\) is positive or negative), after using Pythagorean Theorem, by considering the quadrant.
Use Pythagorean Theorem.
\[x^2+(0.823x)^2~=~1^2\]
Distribute the power of 2 across the two factors in the parentheses. Also, \(1^2=1\).
\[x^2+0.677329x^2~=~1\]
Combine like terms. \(1x^2+0.677329x^2\equiv (1+0.677329)x^2 \equiv 1.677329x^2\)
\[1.677329x^2~=~1\]
Divide both sides by 1.677329.
\[x^2~=~\frac{1}{1.677329}\]
Square root both sides.
\[|x|~=~ \sqrt{\frac{1}{1.677329}}\]
We know the quadrant is quadrant III, so \(x\) is negative.
\[x~=~ -0.7721308\]
We know that \(y=0.823x\).
\[y~=~(0.823)(-0.7721308)\]
\[y~=~-0.6354636\]
So,
\[\sin(\theta)~=~-0.6354636\]
Question
Given:
\[\tan(\theta)~=~-0.45\]\[\theta~\text{ is in quadrant IV}\]
Find \(\sin(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\tan(\theta)\) represents the slope of the segment connecting \((0,0)\) to \((\cos(\theta),\sin(\theta))\). Draw a sketch. We know the point of interest is in quadrant IV. Draw the unit circle centered at the origin and the line \(y=-0.45x\).
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment from the \(x\)-axis to the point. In order for the slope to equal \(\frac{y}{x}\), we can enforce that \(|y|=|-0.45x|\). Notice I am labeling the lengths with positive numbers; I will take care of signs (whether \(x\) or \(y\) is positive or negative), after using Pythagorean Theorem, by considering the quadrant.
Use Pythagorean Theorem.
\[x^2+(0.45x)^2~=~1^2\]
Distribute the power of 2 across the two factors in the parentheses. Also, \(1^2=1\).
\[x^2+0.2025x^2~=~1\]
Combine like terms. \(1x^2+0.2025x^2\equiv (1+0.2025)x^2 \equiv 1.2025x^2\)
\[1.2025x^2~=~1\]
Divide both sides by 1.2025.
\[x^2~=~\frac{1}{1.2025}\]
Square root both sides.
\[|x|~=~ \sqrt{\frac{1}{1.2025}}\]
We know the quadrant is quadrant IV, so \(x\) is positive.
\[x~=~ 0.9119215\]
We know that \(y=-0.45x\).
\[y~=~(-0.45)(0.9119215)\]
\[y~=~-0.4103647\]
So,
\[\sin(\theta)~=~-0.4103647\]
Question
Given:
\[\tan(\theta)~=~-0.196\]\[\theta~\text{ is in quadrant IV}\]
Find \(\sin(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\tan(\theta)\) represents the slope of the segment connecting \((0,0)\) to \((\cos(\theta),\sin(\theta))\). Draw a sketch. We know the point of interest is in quadrant IV. Draw the unit circle centered at the origin and the line \(y=-0.196x\).
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment from the \(x\)-axis to the point. In order for the slope to equal \(\frac{y}{x}\), we can enforce that \(|y|=|-0.196x|\). Notice I am labeling the lengths with positive numbers; I will take care of signs (whether \(x\) or \(y\) is positive or negative), after using Pythagorean Theorem, by considering the quadrant.
Use Pythagorean Theorem.
\[x^2+(0.196x)^2~=~1^2\]
Distribute the power of 2 across the two factors in the parentheses. Also, \(1^2=1\).
\[x^2+0.038416x^2~=~1\]
Combine like terms. \(1x^2+0.038416x^2\equiv (1+0.038416)x^2 \equiv 1.038416x^2\)
\[1.038416x^2~=~1\]
Divide both sides by 1.038416.
\[x^2~=~\frac{1}{1.038416}\]
Square root both sides.
\[|x|~=~ \sqrt{\frac{1}{1.038416}}\]
We know the quadrant is quadrant IV, so \(x\) is positive.
\[x~=~ 0.9813283\]
We know that \(y=-0.196x\).
\[y~=~(-0.196)(0.9813283)\]
\[y~=~-0.1923403\]
So,
\[\sin(\theta)~=~-0.1923403\]
Question
Given:
\[\sin(\theta)~=~-0.241\]\[\theta~\text{ is in quadrant IV}\]
Find \(\cos(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\sin(\theta)\) represents the \(y\) value on the unit circle at an angle of \(\theta\). Draw a sketch. We know the point of interest is in quadrant IV.
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment.
Use Pythagorean Theorem.
\[x^2+0.241^2~=~1^2\]\[x^2+0.241^2~=~1\]
Subtract \(0.241^2\) from both sides.
\[x^2~=~1-0.241^2\]
Square root both sides.
\[|x|~=~\sqrt{(1-0.241^2)}\]
We know the quadrant is quadrant IV, so \(x\) is positive.
\[x~=~0.9705251\]
So,
\[\cos(\theta)~=~0.9705251\]
Question
Given:
\[\sin(\theta)~=~-0.456\]\[\theta~\text{ is in quadrant IV}\]
Find \(\cos(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\sin(\theta)\) represents the \(y\) value on the unit circle at an angle of \(\theta\). Draw a sketch. We know the point of interest is in quadrant IV.
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment.
Use Pythagorean Theorem.
\[x^2+0.456^2~=~1^2\]\[x^2+0.456^2~=~1\]
Subtract \(0.456^2\) from both sides.
\[x^2~=~1-0.456^2\]
Square root both sides.
\[|x|~=~\sqrt{(1-0.456^2)}\]
We know the quadrant is quadrant IV, so \(x\) is positive.
\[x~=~0.8899798\]
So,
\[\cos(\theta)~=~0.8899798\]
Question
Given:
\[\sin(\theta)~=~0.825\]\[\theta~\text{ is in quadrant I}\]
Find \(\cos(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\sin(\theta)\) represents the \(y\) value on the unit circle at an angle of \(\theta\). Draw a sketch. We know the point of interest is in quadrant I.
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment.
Use Pythagorean Theorem.
\[x^2+0.825^2~=~1^2\]\[x^2+0.825^2~=~1\]
Subtract \(0.825^2\) from both sides.
\[x^2~=~1-0.825^2\]
Square root both sides.
\[|x|~=~\sqrt{(1-0.825^2)}\]
We know the quadrant is quadrant I, so \(x\) is positive.
\[x~=~0.5651327\]
So,
\[\cos(\theta)~=~0.5651327\]
Question
Given:
\[\sin(\theta)~=~0.636\]\[\theta~\text{ is in quadrant I}\]
Find \(\cos(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\sin(\theta)\) represents the \(y\) value on the unit circle at an angle of \(\theta\). Draw a sketch. We know the point of interest is in quadrant I.
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment.
Use Pythagorean Theorem.
\[x^2+0.636^2~=~1^2\]\[x^2+0.636^2~=~1\]
Subtract \(0.636^2\) from both sides.
\[x^2~=~1-0.636^2\]
Square root both sides.
\[|x|~=~\sqrt{(1-0.636^2)}\]
We know the quadrant is quadrant I, so \(x\) is positive.
\[x~=~0.7716891\]
So,
\[\cos(\theta)~=~0.7716891\]
Question
Given:
\[\sin(\theta)~=~0.973\]\[\theta~\text{ is in quadrant I}\]
Find \(\cos(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\sin(\theta)\) represents the \(y\) value on the unit circle at an angle of \(\theta\). Draw a sketch. We know the point of interest is in quadrant I.
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment.
Use Pythagorean Theorem.
\[x^2+0.973^2~=~1^2\]\[x^2+0.973^2~=~1\]
Subtract \(0.973^2\) from both sides.
\[x^2~=~1-0.973^2\]
Square root both sides.
\[|x|~=~\sqrt{(1-0.973^2)}\]
We know the quadrant is quadrant I, so \(x\) is positive.
\[x~=~0.2308051\]
So,
\[\cos(\theta)~=~0.2308051\]
Question
Given:
\[\tan(\theta)~=~1.82\]\[\theta~\text{ is in quadrant III}\]
Find \(\cos(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\tan(\theta)\) represents the slope of the segment connecting \((0,0)\) to \((\cos(\theta),\sin(\theta))\). Draw a sketch. We know the point of interest is in quadrant III. Draw the unit circle centered at the origin and the line \(y=1.82x\).
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment from the \(x\)-axis to the point. In order for the slope to equal \(\frac{y}{x}\), we can enforce that \(|y|=|1.82x|\). Notice I am labeling the lengths with positive numbers; I will take care of signs (whether \(x\) or \(y\) is positive or negative), after using Pythagorean Theorem, by considering the quadrant.
Use Pythagorean Theorem.
\[x^2+(1.82x)^2~=~1^2\]
Distribute the power of 2 across the two factors in the parentheses. Also, \(1^2=1\).
\[x^2+3.3124x^2~=~1\]
Combine like terms. \(1x^2+3.3124x^2\equiv (1+3.3124)x^2 \equiv 4.3124x^2\)
\[4.3124x^2~=~1\]
Divide both sides by 4.3124.
\[x^2~=~\frac{1}{4.3124}\]
Square root both sides.
\[|x|~=~ \sqrt{\frac{1}{4.3124}}\]
We know the quadrant is quadrant III, so \(x\) is negative.
\[x~=~ -0.481549\]
So,
\[\cos(\theta)~=~-0.481549\]
Question
Given:
\[\tan(\theta)~=~-3.13\]\[\theta~\text{ is in quadrant II}\]
Find \(\cos(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\tan(\theta)\) represents the slope of the segment connecting \((0,0)\) to \((\cos(\theta),\sin(\theta))\). Draw a sketch. We know the point of interest is in quadrant II. Draw the unit circle centered at the origin and the line \(y=-3.13x\).
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment from the \(x\)-axis to the point. In order for the slope to equal \(\frac{y}{x}\), we can enforce that \(|y|=|-3.13x|\). Notice I am labeling the lengths with positive numbers; I will take care of signs (whether \(x\) or \(y\) is positive or negative), after using Pythagorean Theorem, by considering the quadrant.
Use Pythagorean Theorem.
\[x^2+(3.13x)^2~=~1^2\]
Distribute the power of 2 across the two factors in the parentheses. Also, \(1^2=1\).
\[x^2+9.7969x^2~=~1\]
Combine like terms. \(1x^2+9.7969x^2\equiv (1+9.7969)x^2 \equiv 10.7969x^2\)
\[10.7969x^2~=~1\]
Divide both sides by 10.7969.
\[x^2~=~\frac{1}{10.7969}\]
Square root both sides.
\[|x|~=~ \sqrt{\frac{1}{10.7969}}\]
We know the quadrant is quadrant II, so \(x\) is negative.
\[x~=~ -0.304334\]
So,
\[\cos(\theta)~=~-0.304334\]
Question
Given:
\[\tan(\theta)~=~-1.17\]\[\theta~\text{ is in quadrant IV}\]
Find \(\cos(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\tan(\theta)\) represents the slope of the segment connecting \((0,0)\) to \((\cos(\theta),\sin(\theta))\). Draw a sketch. We know the point of interest is in quadrant IV. Draw the unit circle centered at the origin and the line \(y=-1.17x\).
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment from the \(x\)-axis to the point. In order for the slope to equal \(\frac{y}{x}\), we can enforce that \(|y|=|-1.17x|\). Notice I am labeling the lengths with positive numbers; I will take care of signs (whether \(x\) or \(y\) is positive or negative), after using Pythagorean Theorem, by considering the quadrant.
Use Pythagorean Theorem.
\[x^2+(1.17x)^2~=~1^2\]
Distribute the power of 2 across the two factors in the parentheses. Also, \(1^2=1\).
\[x^2+1.3689x^2~=~1\]
Combine like terms. \(1x^2+1.3689x^2\equiv (1+1.3689)x^2 \equiv 2.3689x^2\)
\[2.3689x^2~=~1\]
Divide both sides by 2.3689.
\[x^2~=~\frac{1}{2.3689}\]
Square root both sides.
\[|x|~=~ \sqrt{\frac{1}{2.3689}}\]
We know the quadrant is quadrant IV, so \(x\) is positive.
\[x~=~ 0.6497206\]
So,
\[\cos(\theta)~=~0.6497206\]
Question
Given:
\[\tan(\theta)~=~-5.22\]\[\theta~\text{ is in quadrant II}\]
Find \(\cos(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\tan(\theta)\) represents the slope of the segment connecting \((0,0)\) to \((\cos(\theta),\sin(\theta))\). Draw a sketch. We know the point of interest is in quadrant II. Draw the unit circle centered at the origin and the line \(y=-5.22x\).
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment from the \(x\)-axis to the point. In order for the slope to equal \(\frac{y}{x}\), we can enforce that \(|y|=|-5.22x|\). Notice I am labeling the lengths with positive numbers; I will take care of signs (whether \(x\) or \(y\) is positive or negative), after using Pythagorean Theorem, by considering the quadrant.
Use Pythagorean Theorem.
\[x^2+(5.22x)^2~=~1^2\]
Distribute the power of 2 across the two factors in the parentheses. Also, \(1^2=1\).
\[x^2+27.2484x^2~=~1\]
Combine like terms. \(1x^2+27.2484x^2\equiv (1+27.2484)x^2 \equiv 28.2484x^2\)
\[28.2484x^2~=~1\]
Divide both sides by 28.2484.
\[x^2~=~\frac{1}{28.2484}\]
Square root both sides.
\[|x|~=~ \sqrt{\frac{1}{28.2484}}\]
We know the quadrant is quadrant II, so \(x\) is negative.
\[x~=~ -0.1881495\]
So,
\[\cos(\theta)~=~-0.1881495\]
Question
Given:
\[\tan(\theta)~=~-1.22\]\[\theta~\text{ is in quadrant IV}\]
Find \(\cos(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\tan(\theta)\) represents the slope of the segment connecting \((0,0)\) to \((\cos(\theta),\sin(\theta))\). Draw a sketch. We know the point of interest is in quadrant IV. Draw the unit circle centered at the origin and the line \(y=-1.22x\).
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment from the \(x\)-axis to the point. In order for the slope to equal \(\frac{y}{x}\), we can enforce that \(|y|=|-1.22x|\). Notice I am labeling the lengths with positive numbers; I will take care of signs (whether \(x\) or \(y\) is positive or negative), after using Pythagorean Theorem, by considering the quadrant.
Use Pythagorean Theorem.
\[x^2+(1.22x)^2~=~1^2\]
Distribute the power of 2 across the two factors in the parentheses. Also, \(1^2=1\).
\[x^2+1.4884x^2~=~1\]
Combine like terms. \(1x^2+1.4884x^2\equiv (1+1.4884)x^2 \equiv 2.4884x^2\)
\[2.4884x^2~=~1\]
Divide both sides by 2.4884.
\[x^2~=~\frac{1}{2.4884}\]
Square root both sides.
\[|x|~=~ \sqrt{\frac{1}{2.4884}}\]
We know the quadrant is quadrant IV, so \(x\) is positive.
\[x~=~ 0.633928\]
So,
\[\cos(\theta)~=~0.633928\]
Question
Given:
\[\sin(\theta)~=~0.987\]\[\theta~\text{ is in quadrant II}\]
Find \(\tan(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\sin(\theta)\) represents the \(y\) value on the unit circle at an angle of \(\theta\). Draw a sketch. We know the point of interest is in quadrant II.
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment.
Use Pythagorean Theorem.
\[x^2+0.987^2~=~1^2\]\[x^2+0.987^2~=~1\]
Subtract \(0.987^2\) from both sides.
\[x^2~=~1-0.987^2\]
Square root both sides.
\[|x|~=~\sqrt{(1-0.987^2)}\]
We know the quadrant is quadrant II, so \(x\) is negative.
\[x~=~-0.1607203\]
So,
\[\cos(\theta)~=~-0.1607203\]
Remember, \(\tan(\theta)\equiv\frac{\sin(\theta)}{\cos(\theta)}\) for any value of \(\theta\). In other words, \(\tan(\theta)\) is the slope of the line connecting the point to the origin.
\[\sin(\theta)~=~-0.27\]\[\theta~\text{ is in quadrant IV}\]
Find \(\tan(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\sin(\theta)\) represents the \(y\) value on the unit circle at an angle of \(\theta\). Draw a sketch. We know the point of interest is in quadrant IV.
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment.
Use Pythagorean Theorem.
\[x^2+0.27^2~=~1^2\]\[x^2+0.27^2~=~1\]
Subtract \(0.27^2\) from both sides.
\[x^2~=~1-0.27^2\]
Square root both sides.
\[|x|~=~\sqrt{(1-0.27^2)}\]
We know the quadrant is quadrant IV, so \(x\) is positive.
\[x~=~0.9628603\]
So,
\[\cos(\theta)~=~0.9628603\]
Remember, \(\tan(\theta)\equiv\frac{\sin(\theta)}{\cos(\theta)}\) for any value of \(\theta\). In other words, \(\tan(\theta)\) is the slope of the line connecting the point to the origin.
\[\sin(\theta)~=~0.468\]\[\theta~\text{ is in quadrant I}\]
Find \(\tan(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\sin(\theta)\) represents the \(y\) value on the unit circle at an angle of \(\theta\). Draw a sketch. We know the point of interest is in quadrant I.
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment.
Use Pythagorean Theorem.
\[x^2+0.468^2~=~1^2\]\[x^2+0.468^2~=~1\]
Subtract \(0.468^2\) from both sides.
\[x^2~=~1-0.468^2\]
Square root both sides.
\[|x|~=~\sqrt{(1-0.468^2)}\]
We know the quadrant is quadrant I, so \(x\) is positive.
\[x~=~0.8837285\]
So,
\[\cos(\theta)~=~0.8837285\]
Remember, \(\tan(\theta)\equiv\frac{\sin(\theta)}{\cos(\theta)}\) for any value of \(\theta\). In other words, \(\tan(\theta)\) is the slope of the line connecting the point to the origin.
\[\sin(\theta)~=~-0.547\]\[\theta~\text{ is in quadrant III}\]
Find \(\tan(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\sin(\theta)\) represents the \(y\) value on the unit circle at an angle of \(\theta\). Draw a sketch. We know the point of interest is in quadrant III.
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment.
Use Pythagorean Theorem.
\[x^2+0.547^2~=~1^2\]\[x^2+0.547^2~=~1\]
Subtract \(0.547^2\) from both sides.
\[x^2~=~1-0.547^2\]
Square root both sides.
\[|x|~=~\sqrt{(1-0.547^2)}\]
We know the quadrant is quadrant III, so \(x\) is negative.
\[x~=~-0.8371326\]
So,
\[\cos(\theta)~=~-0.8371326\]
Remember, \(\tan(\theta)\equiv\frac{\sin(\theta)}{\cos(\theta)}\) for any value of \(\theta\). In other words, \(\tan(\theta)\) is the slope of the line connecting the point to the origin.
\[\sin(\theta)~=~0.765\]\[\theta~\text{ is in quadrant II}\]
Find \(\tan(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\sin(\theta)\) represents the \(y\) value on the unit circle at an angle of \(\theta\). Draw a sketch. We know the point of interest is in quadrant II.
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment.
Use Pythagorean Theorem.
\[x^2+0.765^2~=~1^2\]\[x^2+0.765^2~=~1\]
Subtract \(0.765^2\) from both sides.
\[x^2~=~1-0.765^2\]
Square root both sides.
\[|x|~=~\sqrt{(1-0.765^2)}\]
We know the quadrant is quadrant II, so \(x\) is negative.
\[x~=~-0.6440303\]
So,
\[\cos(\theta)~=~-0.6440303\]
Remember, \(\tan(\theta)\equiv\frac{\sin(\theta)}{\cos(\theta)}\) for any value of \(\theta\). In other words, \(\tan(\theta)\) is the slope of the line connecting the point to the origin.
\[\cos(\theta)~=~0.48\]\[\theta~\text{ is in quadrant I}\]
Find \(\tan(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\cos(\theta)\) represents the \(x\) value on the unit circle at an angle of \(\theta\). Draw a sketch. We know the point of interest is in quadrant I.
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment at \(x=0.48\).
Use Pythagorean Theorem.
\[(0.48)^2+|y|^2~=~1^2\]
The square of a real number always equals the square of the absolute value of the same real number. Also, \(1^2=1\).
\[0.48^2+y^2~=~1\]
Subtract \(0.48^2\) from both sides.
\[y^2~=~1-0.48^2\]
Square root both sides.
\[y~=~\pm \sqrt{(1-0.48^2)}\]
We know the quadrant is quadrant I, so \(y\) is positive.
\[y~=~0.8772685\]
So,
\[\sin(\theta)~=~0.8772685\]
Remember, \(\tan(\theta)\equiv\frac{\sin(\theta)}{\cos(\theta)}\) for any value of \(\theta\). In other words, \(\tan(\theta)\) is the slope of the line connecting the point to the origin.
\[\cos(\theta)~=~0.89\]\[\theta~\text{ is in quadrant IV}\]
Find \(\tan(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\cos(\theta)\) represents the \(x\) value on the unit circle at an angle of \(\theta\). Draw a sketch. We know the point of interest is in quadrant IV.
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment at \(x=0.89\).
Use Pythagorean Theorem.
\[(0.89)^2+|y|^2~=~1^2\]
The square of a real number always equals the square of the absolute value of the same real number. Also, \(1^2=1\).
\[0.89^2+y^2~=~1\]
Subtract \(0.89^2\) from both sides.
\[y^2~=~1-0.89^2\]
Square root both sides.
\[y~=~\pm \sqrt{(1-0.89^2)}\]
We know the quadrant is quadrant IV, so \(y\) is negative.
\[y~=~-0.4559605\]
So,
\[\sin(\theta)~=~-0.4559605\]
Remember, \(\tan(\theta)\equiv\frac{\sin(\theta)}{\cos(\theta)}\) for any value of \(\theta\). In other words, \(\tan(\theta)\) is the slope of the line connecting the point to the origin.
\[\cos(\theta)~=~-0.872\]\[\theta~\text{ is in quadrant II}\]
Find \(\tan(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\cos(\theta)\) represents the \(x\) value on the unit circle at an angle of \(\theta\). Draw a sketch. We know the point of interest is in quadrant II.
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment at \(x=-0.872\).
Use Pythagorean Theorem.
\[(-0.872)^2+|y|^2~=~1^2\]
The square of a real number always equals the square of the absolute value of the same real number. Also, \(1^2=1\).
\[0.872^2+y^2~=~1\]
Subtract \(0.872^2\) from both sides.
\[y^2~=~1-0.872^2\]
Square root both sides.
\[y~=~\pm \sqrt{(1-0.872^2)}\]
We know the quadrant is quadrant II, so \(y\) is positive.
\[y~=~0.4895059\]
So,
\[\sin(\theta)~=~0.4895059\]
Remember, \(\tan(\theta)\equiv\frac{\sin(\theta)}{\cos(\theta)}\) for any value of \(\theta\). In other words, \(\tan(\theta)\) is the slope of the line connecting the point to the origin.
\[\cos(\theta)~=~-0.874\]\[\theta~\text{ is in quadrant III}\]
Find \(\tan(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\cos(\theta)\) represents the \(x\) value on the unit circle at an angle of \(\theta\). Draw a sketch. We know the point of interest is in quadrant III.
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment at \(x=-0.874\).
Use Pythagorean Theorem.
\[(-0.874)^2+|y|^2~=~1^2\]
The square of a real number always equals the square of the absolute value of the same real number. Also, \(1^2=1\).
\[0.874^2+y^2~=~1\]
Subtract \(0.874^2\) from both sides.
\[y^2~=~1-0.874^2\]
Square root both sides.
\[y~=~\pm \sqrt{(1-0.874^2)}\]
We know the quadrant is quadrant III, so \(y\) is negative.
\[y~=~-0.4859259\]
So,
\[\sin(\theta)~=~-0.4859259\]
Remember, \(\tan(\theta)\equiv\frac{\sin(\theta)}{\cos(\theta)}\) for any value of \(\theta\). In other words, \(\tan(\theta)\) is the slope of the line connecting the point to the origin.
\[\cos(\theta)~=~0.264\]\[\theta~\text{ is in quadrant IV}\]
Find \(\tan(\theta)\). The tolerance is \(\pm 0.001\).
Solution
Remember, \(\cos(\theta)\) represents the \(x\) value on the unit circle at an angle of \(\theta\). Draw a sketch. We know the point of interest is in quadrant IV.
You can draw a right triangle. The standard way to do this is by using a segment along the \(x\)-axis and a vertical segment at \(x=0.264\).
Use Pythagorean Theorem.
\[(0.264)^2+|y|^2~=~1^2\]
The square of a real number always equals the square of the absolute value of the same real number. Also, \(1^2=1\).
\[0.264^2+y^2~=~1\]
Subtract \(0.264^2\) from both sides.
\[y^2~=~1-0.264^2\]
Square root both sides.
\[y~=~\pm \sqrt{(1-0.264^2)}\]
We know the quadrant is quadrant IV, so \(y\) is negative.
\[y~=~-0.9645227\]
So,
\[\sin(\theta)~=~-0.9645227\]
Remember, \(\tan(\theta)\equiv\frac{\sin(\theta)}{\cos(\theta)}\) for any value of \(\theta\). In other words, \(\tan(\theta)\) is the slope of the line connecting the point to the origin.
At noon, on an equinox, at the equator, the sun is directly overhead. Over land with minimal elevation change, we can approximate the local ground as flat.
A rocket is launched at an angle of elevation of 0.643 radians above the horizon at a speed of 3630 meters per second.
How fast, in meters per second, is the shadow moving across the flat ground? The tolerance is \(\pm 10\) m/s.
Solution
We recognize that velocity is a vector. In this scenario, we can think of the vector as a 2D vector given in polar form.
Using basic right-triangle trigonometry, you can find the horizontal component of the velocity.
Draw a vector with magnitude (length) of 3630 m/s and angle of 0.643 radians. We can indicate the magnitude with double vertical lines on either side of the vector’s name. We can also indicate magnitude by leaving off the arrow.
\[||\vec{v}||~=~ v ~=~3630~\mathrm{\frac{m}{s}}\]
We want to decompose the vector into its horizontal component (\(\vec{v}_x\)) and its vertical component (\(\vec{v}_y\)). The horizontal component’s magnitude (denoted as \(||\vec{v}_x||\), or more simply as \(v_x\)) is the speed of the shadow.
We can set up the cosine equation.
\[\cos(0.643)~=~\frac{v_x}{3630}\]
Solve for \(v_x\).
\[v_x~=~3630\cdot\cos(0.643)\]
Evaluate with a calculator. Remember, we are in RADIAN MODE!
\[v_x~=~2905.0910503\]\[v_x~\approx~2910\]
Question
At noon, on an equinox, at the equator, the sun is directly overhead. Over land with minimal elevation change, we can approximate the local ground as flat.
A rocket is launched at an angle of elevation of 0.889 radians above the horizon at a speed of 2810 meters per second.
How fast, in meters per second, is the shadow moving across the flat ground? The tolerance is \(\pm 10\) m/s.
Solution
We recognize that velocity is a vector. In this scenario, we can think of the vector as a 2D vector given in polar form.
Using basic right-triangle trigonometry, you can find the horizontal component of the velocity.
Draw a vector with magnitude (length) of 2810 m/s and angle of 0.889 radians. We can indicate the magnitude with double vertical lines on either side of the vector’s name. We can also indicate magnitude by leaving off the arrow.
\[||\vec{v}||~=~ v ~=~2810~\mathrm{\frac{m}{s}}\]
We want to decompose the vector into its horizontal component (\(\vec{v}_x\)) and its vertical component (\(\vec{v}_y\)). The horizontal component’s magnitude (denoted as \(||\vec{v}_x||\), or more simply as \(v_x\)) is the speed of the shadow.
We can set up the cosine equation.
\[\cos(0.889)~=~\frac{v_x}{2810}\]
Solve for \(v_x\).
\[v_x~=~2810\cdot\cos(0.889)\]
Evaluate with a calculator. Remember, we are in RADIAN MODE!
\[v_x~=~1770.8304816\]\[v_x~\approx~1770\]
Question
At noon, on an equinox, at the equator, the sun is directly overhead. Over land with minimal elevation change, we can approximate the local ground as flat.
A rocket is launched at an angle of elevation of 0.508 radians above the horizon at a speed of 3140 meters per second.
How fast, in meters per second, is the shadow moving across the flat ground? The tolerance is \(\pm 10\) m/s.
Solution
We recognize that velocity is a vector. In this scenario, we can think of the vector as a 2D vector given in polar form.
Using basic right-triangle trigonometry, you can find the horizontal component of the velocity.
Draw a vector with magnitude (length) of 3140 m/s and angle of 0.508 radians. We can indicate the magnitude with double vertical lines on either side of the vector’s name. We can also indicate magnitude by leaving off the arrow.
\[||\vec{v}||~=~ v ~=~3140~\mathrm{\frac{m}{s}}\]
We want to decompose the vector into its horizontal component (\(\vec{v}_x\)) and its vertical component (\(\vec{v}_y\)). The horizontal component’s magnitude (denoted as \(||\vec{v}_x||\), or more simply as \(v_x\)) is the speed of the shadow.
We can set up the cosine equation.
\[\cos(0.508)~=~\frac{v_x}{3140}\]
Solve for \(v_x\).
\[v_x~=~3140\cdot\cos(0.508)\]
Evaluate with a calculator. Remember, we are in RADIAN MODE!
\[v_x~=~2743.4780242\]\[v_x~\approx~2740\]
Question
At noon, on an equinox, at the equator, the sun is directly overhead. Over land with minimal elevation change, we can approximate the local ground as flat.
A rocket is launched at an angle of elevation of 0.76 radians above the horizon at a speed of 4510 meters per second.
How fast, in meters per second, is the shadow moving across the flat ground? The tolerance is \(\pm 10\) m/s.
Solution
We recognize that velocity is a vector. In this scenario, we can think of the vector as a 2D vector given in polar form.
Using basic right-triangle trigonometry, you can find the horizontal component of the velocity.
Draw a vector with magnitude (length) of 4510 m/s and angle of 0.76 radians. We can indicate the magnitude with double vertical lines on either side of the vector’s name. We can also indicate magnitude by leaving off the arrow.
\[||\vec{v}||~=~ v ~=~4510~\mathrm{\frac{m}{s}}\]
We want to decompose the vector into its horizontal component (\(\vec{v}_x\)) and its vertical component (\(\vec{v}_y\)). The horizontal component’s magnitude (denoted as \(||\vec{v}_x||\), or more simply as \(v_x\)) is the speed of the shadow.
We can set up the cosine equation.
\[\cos(0.76)~=~\frac{v_x}{4510}\]
Solve for \(v_x\).
\[v_x~=~4510\cdot\cos(0.76)\]
Evaluate with a calculator. Remember, we are in RADIAN MODE!
\[v_x~=~3269.0104084\]\[v_x~\approx~3270\]
Question
At noon, on an equinox, at the equator, the sun is directly overhead. Over land with minimal elevation change, we can approximate the local ground as flat.
A rocket is launched at an angle of elevation of 0.429 radians above the horizon at a speed of 4030 meters per second.
How fast, in meters per second, is the shadow moving across the flat ground? The tolerance is \(\pm 10\) m/s.
Solution
We recognize that velocity is a vector. In this scenario, we can think of the vector as a 2D vector given in polar form.
Using basic right-triangle trigonometry, you can find the horizontal component of the velocity.
Draw a vector with magnitude (length) of 4030 m/s and angle of 0.429 radians. We can indicate the magnitude with double vertical lines on either side of the vector’s name. We can also indicate magnitude by leaving off the arrow.
\[||\vec{v}||~=~ v ~=~4030~\mathrm{\frac{m}{s}}\]
We want to decompose the vector into its horizontal component (\(\vec{v}_x\)) and its vertical component (\(\vec{v}_y\)). The horizontal component’s magnitude (denoted as \(||\vec{v}_x||\), or more simply as \(v_x\)) is the speed of the shadow.
We can set up the cosine equation.
\[\cos(0.429)~=~\frac{v_x}{4030}\]
Solve for \(v_x\).
\[v_x~=~4030\cdot\cos(0.429)\]
Evaluate with a calculator. Remember, we are in RADIAN MODE!
\[v_x~=~3664.8101287\]\[v_x~\approx~3660\]
Question
At sunset, the sun’s rays are parallel to the local flat ground. A rocket is launched at an angle of elevation of 0.782 radians above the horizon at a speed of 4710 meters per second. Its shadow is projected onto a tall cliff to the east.
How fast, in meters per second, is the shadow moving up the cliff? The tolerance is \(\pm 10\) m/s.
Solution
We recognize that velocity is a vector. In this scenario, we can think of the vector as a 2D vector given in polar form.
Using basic right-triangle trigonometry, you can find the horizontal component of the velocity.
Draw a vector with magnitude (length) of 4710 m/s and angle of 0.782 radians. We can indicate the magnitude with double vertical lines on either side of the vector’s name. We can also indicate magnitude by leaving off the arrow.
\[||\vec{v}||~=~ v ~=~4710~\mathrm{\frac{m}{s}}\]
We want to decompose the vector into its horizontal component (\(\vec{v}_x\)) and its vertical component (\(\vec{v}_y\)). The vertical component’s magnitude (denoted as \(||\vec{v}_y||\), or more simply as \(v_y\)) is the speed of the shadow.
We can set up the sine equation.
\[\sin(0.782)~=~\frac{v_y}{4710}\]
Solve for \(v_y\).
\[v_y~=~4710\cdot\sin(0.782)\]
Evaluate with a calculator. Remember, we are in RADIAN MODE!
\[v_y~=~3319.1362406\]\[v_y~\approx~3320\]
Question
At sunset, the sun’s rays are parallel to the local flat ground. A rocket is launched at an angle of elevation of 0.834 radians above the horizon at a speed of 2390 meters per second. Its shadow is projected onto a tall cliff to the east.
How fast, in meters per second, is the shadow moving up the cliff? The tolerance is \(\pm 10\) m/s.
Solution
We recognize that velocity is a vector. In this scenario, we can think of the vector as a 2D vector given in polar form.
Using basic right-triangle trigonometry, you can find the horizontal component of the velocity.
Draw a vector with magnitude (length) of 2390 m/s and angle of 0.834 radians. We can indicate the magnitude with double vertical lines on either side of the vector’s name. We can also indicate magnitude by leaving off the arrow.
\[||\vec{v}||~=~ v ~=~2390~\mathrm{\frac{m}{s}}\]
We want to decompose the vector into its horizontal component (\(\vec{v}_x\)) and its vertical component (\(\vec{v}_y\)). The vertical component’s magnitude (denoted as \(||\vec{v}_y||\), or more simply as \(v_y\)) is the speed of the shadow.
We can set up the sine equation.
\[\sin(0.834)~=~\frac{v_y}{2390}\]
Solve for \(v_y\).
\[v_y~=~2390\cdot\sin(0.834)\]
Evaluate with a calculator. Remember, we are in RADIAN MODE!
\[v_y~=~1770.0936628\]\[v_y~\approx~1770\]
Question
At sunset, the sun’s rays are parallel to the local flat ground. A rocket is launched at an angle of elevation of 0.731 radians above the horizon at a speed of 2580 meters per second. Its shadow is projected onto a tall cliff to the east.
How fast, in meters per second, is the shadow moving up the cliff? The tolerance is \(\pm 10\) m/s.
Solution
We recognize that velocity is a vector. In this scenario, we can think of the vector as a 2D vector given in polar form.
Using basic right-triangle trigonometry, you can find the horizontal component of the velocity.
Draw a vector with magnitude (length) of 2580 m/s and angle of 0.731 radians. We can indicate the magnitude with double vertical lines on either side of the vector’s name. We can also indicate magnitude by leaving off the arrow.
\[||\vec{v}||~=~ v ~=~2580~\mathrm{\frac{m}{s}}\]
We want to decompose the vector into its horizontal component (\(\vec{v}_x\)) and its vertical component (\(\vec{v}_y\)). The vertical component’s magnitude (denoted as \(||\vec{v}_y||\), or more simply as \(v_y\)) is the speed of the shadow.
We can set up the sine equation.
\[\sin(0.731)~=~\frac{v_y}{2580}\]
Solve for \(v_y\).
\[v_y~=~2580\cdot\sin(0.731)\]
Evaluate with a calculator. Remember, we are in RADIAN MODE!
\[v_y~=~1722.4453477\]\[v_y~\approx~1720\]
Question
At sunset, the sun’s rays are parallel to the local flat ground. A rocket is launched at an angle of elevation of 0.724 radians above the horizon at a speed of 2680 meters per second. Its shadow is projected onto a tall cliff to the east.
How fast, in meters per second, is the shadow moving up the cliff? The tolerance is \(\pm 10\) m/s.
Solution
We recognize that velocity is a vector. In this scenario, we can think of the vector as a 2D vector given in polar form.
Using basic right-triangle trigonometry, you can find the horizontal component of the velocity.
Draw a vector with magnitude (length) of 2680 m/s and angle of 0.724 radians. We can indicate the magnitude with double vertical lines on either side of the vector’s name. We can also indicate magnitude by leaving off the arrow.
\[||\vec{v}||~=~ v ~=~2680~\mathrm{\frac{m}{s}}\]
We want to decompose the vector into its horizontal component (\(\vec{v}_x\)) and its vertical component (\(\vec{v}_y\)). The vertical component’s magnitude (denoted as \(||\vec{v}_y||\), or more simply as \(v_y\)) is the speed of the shadow.
We can set up the sine equation.
\[\sin(0.724)~=~\frac{v_y}{2680}\]
Solve for \(v_y\).
\[v_y~=~2680\cdot\sin(0.724)\]
Evaluate with a calculator. Remember, we are in RADIAN MODE!
\[v_y~=~1775.1961196\]\[v_y~\approx~1780\]
Question
At sunset, the sun’s rays are parallel to the local flat ground. A rocket is launched at an angle of elevation of 1.22 radians above the horizon at a speed of 3350 meters per second. Its shadow is projected onto a tall cliff to the east.
How fast, in meters per second, is the shadow moving up the cliff? The tolerance is \(\pm 10\) m/s.
Solution
We recognize that velocity is a vector. In this scenario, we can think of the vector as a 2D vector given in polar form.
Using basic right-triangle trigonometry, you can find the horizontal component of the velocity.
Draw a vector with magnitude (length) of 3350 m/s and angle of 1.22 radians. We can indicate the magnitude with double vertical lines on either side of the vector’s name. We can also indicate magnitude by leaving off the arrow.
\[||\vec{v}||~=~ v ~=~3350~\mathrm{\frac{m}{s}}\]
We want to decompose the vector into its horizontal component (\(\vec{v}_x\)) and its vertical component (\(\vec{v}_y\)). The vertical component’s magnitude (denoted as \(||\vec{v}_y||\), or more simply as \(v_y\)) is the speed of the shadow.
We can set up the sine equation.
\[\sin(1.22)~=~\frac{v_y}{3350}\]
Solve for \(v_y\).
\[v_y~=~3350\cdot\sin(1.22)\]
Evaluate with a calculator. Remember, we are in RADIAN MODE!
\[v_y~=~3145.9828437\]\[v_y~\approx~3150\]
Question
A point on a Cartesian plane is given in polar coordinates:
\[r = 5.07\]\[\theta=\frac{5\pi}{7}\]
Please determine the rectangular coordinates (\(x\) and \(y\)).
The horizontal coordinate: \(x=\)
The vertical coordinate: \(y=\)
Solution
We can use right-triangle trigonometry. Start by finding the reference angle. To do this, find the absolute difference between the angle and the nearest multiple of \(\pi\). So, in this case, the reference angle (\(\phi\)) is:
\[\phi ~=~ \left|\frac{5\pi}{7}-\pi\right|\]
\[\phi ~=~ \frac{2\pi}{7}\]
Remember, \(r=5.07\). Use trigonometry to calculate the lengths of the right triangle.
Since we are in quadrant 2, we know the signs of \(x\) and \(y\).
\[x = -3.1610933\]
\[y = 3.9638856\]
If you round to the hundredths place:
\[x = -3.16\]\[y = 3.96\]
Another approach
It turns out that the trigonometric functions are defined for all angles (not just acute angles). The way they are defined actually makes this problem trivial. There is no need to find an acute reference angle.
\[x = r \cos(\theta)\]
\[y = r \sin(\theta)\]
Using these above formulas will work directly.
Question
A point on a Cartesian plane is given in polar coordinates:
\[r = 4.46\]\[\theta=\frac{4\pi}{5}\]
Please determine the rectangular coordinates (\(x\) and \(y\)).
The horizontal coordinate: \(x=\)
The vertical coordinate: \(y=\)
Solution
We can use right-triangle trigonometry. Start by finding the reference angle. To do this, find the absolute difference between the angle and the nearest multiple of \(\pi\). So, in this case, the reference angle (\(\phi\)) is:
\[\phi ~=~ \left|\frac{4\pi}{5}-\pi\right|\]
\[\phi ~=~ \frac{\pi}{5}\]
Remember, \(r=4.46\). Use trigonometry to calculate the lengths of the right triangle.
Since we are in quadrant 2, we know the signs of \(x\) and \(y\).
\[x = -3.6082158\]
\[y = 2.6215222\]
If you round to the hundredths place:
\[x = -3.61\]\[y = 2.62\]
Another approach
It turns out that the trigonometric functions are defined for all angles (not just acute angles). The way they are defined actually makes this problem trivial. There is no need to find an acute reference angle.
\[x = r \cos(\theta)\]
\[y = r \sin(\theta)\]
Using these above formulas will work directly.
Question
A point on a Cartesian plane is given in polar coordinates:
\[r = 3.75\]\[\theta=\frac{17\pi}{15}\]
Please determine the rectangular coordinates (\(x\) and \(y\)).
The horizontal coordinate: \(x=\)
The vertical coordinate: \(y=\)
Solution
We can use right-triangle trigonometry. Start by finding the reference angle. To do this, find the absolute difference between the angle and the nearest multiple of \(\pi\). So, in this case, the reference angle (\(\phi\)) is:
\[\phi ~=~ \left|\frac{17\pi}{15}-\pi\right|\]
\[\phi ~=~ \frac{2\pi}{15}\]
Remember, \(r=3.75\). Use trigonometry to calculate the lengths of the right triangle.
Since we are in quadrant 3, we know the signs of \(x\) and \(y\).
\[x = -3.4257955\]
\[y = -1.5252624\]
If you round to the hundredths place:
\[x = -3.43\]\[y = -1.53\]
Another approach
It turns out that the trigonometric functions are defined for all angles (not just acute angles). The way they are defined actually makes this problem trivial. There is no need to find an acute reference angle.
\[x = r \cos(\theta)\]
\[y = r \sin(\theta)\]
Using these above formulas will work directly.
Question
A point on a Cartesian plane is given in polar coordinates:
\[r = 4.42\]\[\theta=\frac{9\pi}{5}\]
Please determine the rectangular coordinates (\(x\) and \(y\)).
The horizontal coordinate: \(x=\)
The vertical coordinate: \(y=\)
Solution
We can use right-triangle trigonometry. Start by finding the reference angle. To do this, find the absolute difference between the angle and the nearest multiple of \(\pi\). So, in this case, the reference angle (\(\phi\)) is:
\[\phi ~=~ \left|\frac{9\pi}{5}-2\pi\right|\]
\[\phi ~=~ \frac{\pi}{5}\]
Remember, \(r=4.42\). Use trigonometry to calculate the lengths of the right triangle.
Since we are in quadrant 4, we know the signs of \(x\) and \(y\).
\[x = 3.5758551\]
\[y = -2.5980108\]
If you round to the hundredths place:
\[x = 3.58\]\[y = -2.6\]
Another approach
It turns out that the trigonometric functions are defined for all angles (not just acute angles). The way they are defined actually makes this problem trivial. There is no need to find an acute reference angle.
\[x = r \cos(\theta)\]
\[y = r \sin(\theta)\]
Using these above formulas will work directly.
Question
A point on a Cartesian plane is given in polar coordinates:
\[r = 3.61\]\[\theta=\frac{9\pi}{8}\]
Please determine the rectangular coordinates (\(x\) and \(y\)).
The horizontal coordinate: \(x=\)
The vertical coordinate: \(y=\)
Solution
We can use right-triangle trigonometry. Start by finding the reference angle. To do this, find the absolute difference between the angle and the nearest multiple of \(\pi\). So, in this case, the reference angle (\(\phi\)) is:
\[\phi ~=~ \left|\frac{9\pi}{8}-\pi\right|\]
\[\phi ~=~ \frac{\pi}{8}\]
Remember, \(r=3.61\). Use trigonometry to calculate the lengths of the right triangle.
Since we are in quadrant 3, we know the signs of \(x\) and \(y\).
\[x = -3.3352051\]
\[y = -1.3814872\]
If you round to the hundredths place:
\[x = -3.34\]\[y = -1.38\]
Another approach
It turns out that the trigonometric functions are defined for all angles (not just acute angles). The way they are defined actually makes this problem trivial. There is no need to find an acute reference angle.
Because the point is in quadrant II, we find the difference of \(\pi\) minus \(\phi\) to get \(\theta\).
\[\theta = \pi-\phi\]
\[\theta\approx2.3\]
Unfortunately, you cannot simply use \(\arctan\left(\frac{3.13}{-2.82}\right)\) to get the angle \(\theta\). This is because tangent is not a one-to-one function.
Now, Desmos (and many other calculators) allow you to feed \(y\) and \(x\) as two seperate arguments into the arctangent function: \(\arctan(3.13,-2.82)\). This will give you a correct angle, between \(-\pi\) and \(\pi\). If you get a negative angle, you can just add \(2\pi\) to get a coterminal angle between \(0\) and \(2\pi\), as the prompt asks for.
Because the point is in quadrant IV, we find the difference of \(2\pi\) minus \(\phi\) to get \(\theta\).
\[\theta = 2\pi-\phi\]
\[\theta\approx5.8\]
Unfortunately, you cannot simply use \(\arctan\left(\frac{-1.85}{3.52}\right)\) to get the angle \(\theta\). This is because tangent is not a one-to-one function.
Now, Desmos (and many other calculators) allow you to feed \(y\) and \(x\) as two seperate arguments into the arctangent function: \(\arctan(-1.85,3.52)\). This will give you a correct angle, between \(-\pi\) and \(\pi\). If you get a negative angle, you can just add \(2\pi\) to get a coterminal angle between \(0\) and \(2\pi\), as the prompt asks for.
Find the reference angle by using the arctan function with the quotient of the absolute values of the rectangular coordinates.
\[\phi ~=~ \arctan\left(\frac{2.18}{2.6}\right)\]
\[\phi ~=~ 0.6977571\]
Because the point is in quadrant IV, we find the difference of \(2\pi\) minus \(\phi\) to get \(\theta\).
\[\theta = 2\pi-\phi\]
\[\theta\approx5.59\]
Unfortunately, you cannot simply use \(\arctan\left(\frac{-2.18}{2.6}\right)\) to get the angle \(\theta\). This is because tangent is not a one-to-one function.
Now, Desmos (and many other calculators) allow you to feed \(y\) and \(x\) as two seperate arguments into the arctangent function: \(\arctan(-2.18,2.6)\). This will give you a correct angle, between \(-\pi\) and \(\pi\). If you get a negative angle, you can just add \(2\pi\) to get a coterminal angle between \(0\) and \(2\pi\), as the prompt asks for.
Because the point is in quadrant III, we add \(\pi\) and \(\phi\) to get \(\theta\).
\[\theta = \pi+\phi\]
\[\theta\approx3.62\]
Unfortunately, you cannot simply use \(\arctan\left(\frac{-2.03}{-3.87}\right)\) to get the angle \(\theta\). This is because tangent is not a one-to-one function.
Now, Desmos (and many other calculators) allow you to feed \(y\) and \(x\) as two seperate arguments into the arctangent function: \(\arctan(-2.03,-3.87)\). This will give you a correct angle, between \(-\pi\) and \(\pi\). If you get a negative angle, you can just add \(2\pi\) to get a coterminal angle between \(0\) and \(2\pi\), as the prompt asks for.
Because the point is in quadrant IV, we find the difference of \(2\pi\) minus \(\phi\) to get \(\theta\).
\[\theta = 2\pi-\phi\]
\[\theta\approx5.32\]
Unfortunately, you cannot simply use \(\arctan\left(\frac{-3.78}{2.61}\right)\) to get the angle \(\theta\). This is because tangent is not a one-to-one function.
Now, Desmos (and many other calculators) allow you to feed \(y\) and \(x\) as two seperate arguments into the arctangent function: \(\arctan(-3.78,2.61)\). This will give you a correct angle, between \(-\pi\) and \(\pi\). If you get a negative angle, you can just add \(2\pi\) to get a coterminal angle between \(0\) and \(2\pi\), as the prompt asks for.
Question
A canoe travels 2.3 m/s relative to the water. A river, heading east, flows at 3.5 m/s. The canoe points at a direction 1.43 radians counterclockwise from east.
Find the canoe’s total speed, in m/s, and direction of travel, as radians counterclockwise from east. (In other words, find the canoe’s velocity relative to the ground, in polar form.)
We will use \(\vec{v}_1\) for the canoe’s velocity relative to water. It is shown below with approximate rectangular coordinates (found from formulas above, using \(r=2.3\) and \(\theta=1.43\)).
\[\vec{v}_1 = (0.323,2.277)\]
We will use \(\vec{v}_2\) for the river’s velocity vector. Since the river flows directly east, all of its speed is in the positive horizontal direction.
\[\vec{v}_2 = (3.500,0.000)\]
Now, add the corresponding components to find \(\vec{v}_3\), the velocity of the canoe relative to the ground.
So, if we round like the answer choices, the speed of the canoe is 4.4 m/s at a direction of 0.54 radians counterclockwise from east.
Question
A canoe travels 2.8 m/s relative to the water. A river, heading east, flows at 2.3 m/s. The canoe points at a direction 0.58 radians counterclockwise from east.
Find the canoe’s total speed, in m/s, and direction of travel, as radians counterclockwise from east. (In other words, find the canoe’s velocity relative to the ground, in polar form.)
We will use \(\vec{v}_1\) for the canoe’s velocity relative to water. It is shown below with approximate rectangular coordinates (found from formulas above, using \(r=2.8\) and \(\theta=0.58\)).
\[\vec{v}_1 = (2.342,1.534)\]
We will use \(\vec{v}_2\) for the river’s velocity vector. Since the river flows directly east, all of its speed is in the positive horizontal direction.
\[\vec{v}_2 = (2.300,0.000)\]
Now, add the corresponding components to find \(\vec{v}_3\), the velocity of the canoe relative to the ground.
So, if we round like the answer choices, the speed of the canoe is 4.9 m/s at a direction of 0.32 radians counterclockwise from east.
Question
A canoe travels 3.3 m/s relative to the water. A river, heading east, flows at 2.0 m/s. The canoe points at a direction 0.89 radians counterclockwise from east.
Find the canoe’s total speed, in m/s, and direction of travel, as radians counterclockwise from east. (In other words, find the canoe’s velocity relative to the ground, in polar form.)
We will use \(\vec{v}_1\) for the canoe’s velocity relative to water. It is shown below with approximate rectangular coordinates (found from formulas above, using \(r=3.3\) and \(\theta=0.89\)).
\[\vec{v}_1 = (2.077,2.564)\]
We will use \(\vec{v}_2\) for the river’s velocity vector. Since the river flows directly east, all of its speed is in the positive horizontal direction.
\[\vec{v}_2 = (2.000,0.000)\]
Now, add the corresponding components to find \(\vec{v}_3\), the velocity of the canoe relative to the ground.
So, if we round like the answer choices, the speed of the canoe is 4.8 m/s at a direction of 0.56 radians counterclockwise from east.
Question
A canoe travels 1.9 m/s relative to the water. A river, heading east, flows at 2.2 m/s. The canoe points at a direction 0.64 radians counterclockwise from east.
Find the canoe’s total speed, in m/s, and direction of travel, as radians counterclockwise from east. (In other words, find the canoe’s velocity relative to the ground, in polar form.)
We will use \(\vec{v}_1\) for the canoe’s velocity relative to water. It is shown below with approximate rectangular coordinates (found from formulas above, using \(r=1.9\) and \(\theta=0.64\)).
\[\vec{v}_1 = (1.524,1.135)\]
We will use \(\vec{v}_2\) for the river’s velocity vector. Since the river flows directly east, all of its speed is in the positive horizontal direction.
\[\vec{v}_2 = (2.200,0.000)\]
Now, add the corresponding components to find \(\vec{v}_3\), the velocity of the canoe relative to the ground.
So, if we round like the answer choices, the speed of the canoe is 3.9 m/s at a direction of 0.30 radians counterclockwise from east.
Question
A canoe travels 1.7 m/s relative to the water. A river, heading east, flows at 2.8 m/s. The canoe points at a direction 2.08 radians counterclockwise from east.
Find the canoe’s total speed, in m/s, and direction of travel, as radians counterclockwise from east. (In other words, find the canoe’s velocity relative to the ground, in polar form.)
We will use \(\vec{v}_1\) for the canoe’s velocity relative to water. It is shown below with approximate rectangular coordinates (found from formulas above, using \(r=1.7\) and \(\theta=2.08\)).
\[\vec{v}_1 = (-0.829,1.484)\]
We will use \(\vec{v}_2\) for the river’s velocity vector. Since the river flows directly east, all of its speed is in the positive horizontal direction.
\[\vec{v}_2 = (2.800,0.000)\]
Now, add the corresponding components to find \(\vec{v}_3\), the velocity of the canoe relative to the ground.
